Where do titration calculations go wrong?

The mole ratio is forgotten: manganate(VII) to iron(II) is one to five, and using one to one undercounts by five times.

What is the fixed route for electrolysis sums?

Charge is current times time, divide by 96 500 for moles of electrons, then use the electron ratio of the half-equation.

S4 Chemistry: Redox and Electrolysis (Paper 4)

2026-11-01

Tutor Prep Guide · Secondary Chemistry

This is the error-repair and mixed paper: oxidation numbers, titration moles and electrolysis charge together.

Direct answer　Write the oxidation number or the mole value first, then check with the conservation of electrons, atoms or charge.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	25
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: S4 Chemistry: Redox and Electrolysis
Level	Four basic, three advanced, two challenge and one error-repair item

02　Questions (10 items)

FoundationFind the oxidation number of chromium in potassium dichromate, K2Cr2O7.
FoundationFind the oxidation number of sulphur in the sulphate ion, SO42-.
FoundationFind the oxidation number of nitrogen in the ammonium ion, NH4+.
FoundationIs an oxidising agent itself oxidised or reduced?
IntermediateHow many times must the Fe2+ half-equation be multiplied to combine with MnO4- plus 8H+ plus 5e-?
IntermediateA titration uses 24.0 cm3 of 0.0200 mol dm-3 KMnO4. Find the moles of Fe2+.
IntermediateElectrolysis of copper(II) sulphate passes 1.50 A for 772 s. What mass of copper forms at the cathode with 96 500 C per mole of electrons?
ChallengeWhat volume of 0.0200 mol dm-3 KMnO4 oxidises 25.0 cm3 of 0.100 mol dm-3 Fe2+ completely?
ChallengeWith inert electrodes, 2.00 A for 1 930 s electrolyses copper(II) sulphate. Find the mass of copper and the volume of oxygen at 24.0 dm3 per mole.
Error redoError repair: a student writes that a rise in oxidation number is reduction. Correct it with one example.

03　Answers and one-line notes

#	Answer	One-line note
1	+6	Two potassiums give plus two and seven oxygens minus fourteen, so the two chromiums share plus twelve.
2	+6	Four oxygens give minus eight and the ion carries minus two, so sulphur must be plus six.
3	-3	Four hydrogens give plus four and the ion carries plus one, so nitrogen is minus three.
4	reduced	An oxidising agent gains electrons and its oxidation number falls, so it is itself reduced.
5	5	Manganese needs five electrons, so the iron half-equation is multiplied by five.
6	0.00240 mol	The manganate(VII) is 0.0200 times 0.0240, which is 0.000480 mol, and times five gives 0.00240 mol.
7	0.381 g	The charge is 1 158 C, which is 0.0120 mol of electrons; copper needs two electrons, so 0.00600 mol or 0.381 g forms.
8	25.0 cm3	There is 0.00250 mol of Fe2+, so only 0.000500 mol of manganate(VII) is needed, which is 0.0250 dm3.
9	copper 1.27 g; oxygen 0.240 dm3	The charge of 3 860 C gives 0.0400 mol of electrons, so 0.0200 mol of copper or 1.27 g and 0.0100 mol of oxygen or 0.240 dm3.
10	a rise means oxidation, for example zinc going from 0 to +2	Losing electrons raises the oxidation number and is called oxidation, while gaining electrons lowers it and is reduction.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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